91Ó°ÊÓ

- Water flowing at depth is

$\mathrm{h}=10\mathrm{cm}$

$=0.1\mathrm{m}$

- Tank holding water to height is

$\mathrm{H}=40\mathrm{cm}$

$=0.4\mathrm{m}$

By using Bernoulli's equation, we can find the speed $\mathit{v}$of flow of water. Then by using value of $\mathit{v}$, we can find the distance $\mathit{x}$. The value of $\mathit{x}$depends on velocity $\mathit{V}$and timet. If here, $\mathit{x}$is the same, we can take the appropriate value for $\mathit{V}$, and by using the value of $\mathit{V}$, we can find the depth for a second hole to be made to give the same value of $\mathit{x}$. Finally, for ${\mathit{X}}_{\mathbf{m}\mathbf{a}\mathbf{x}}$speed of flow of water

must be maximum. By using this condition, we can find the value of depth for such a hole to be made to give maximum $\mathit{x}$.

Formula:

Bernoulli's equation

role="math" localid="1657691556271" $P+\frac{1}{2}\mathrm{ÒÏ}{V}^2+\mathrm{ÒÏ}gy=\text{constant}$

Kinematic equation

$y=y_0+v_{0}t+\frac{1}{2}a{t}^2$

Velocity

$v=\frac{y}{t}$

The total energy flow for the flow of water just outside the hole is $P+\frac{1}{2}\mathrm{ÒÏ}{V}^2+\mathrm{ÒÏ}g{y}_1=$constant

The total energy flow for the flow of water just inside the hole is $p+\frac{1}{2}\mathrm{ÒÏ}{v}^2+\mathrm{ÒÏ}g{y}_2=$constant

$∴$The energy flow rate of the water in the hole is$P+\frac{1}{2}\mathrm{ÒÏ}{V}^2+\mathrm{ÒÏ}g{y}_1=p+\frac{1}{2}\mathrm{ÒÏ}{v}^2+\mathrm{ÒÏ}g{y}_2=$constant

But the flow of fluid is on the same level form the ground; thus, $y_1=y_2$

and the speed of the water just outside the pipe is equal to zero. Thus, $V=0$

It gives,

$P=p+\frac{1}{2}\mathrm{ÒÏ}{v}^2$

$v=\sqrt{\frac{2(P-p)}{\mathrm{ÒÏ}}}$

The pressure difference between the outside and inside of the hole is equal to the pressure due to the depth $h$

Thus,

$\left(P_{\mathrm{θ}}-P_a\right)=(P-p)$

role="math" localid="1657692258715" $=pgh$

Putting this value, we get

$v=\sqrt{\frac{2\mathrm{ÒÏ}gh}{\mathrm{ÒÏ}}}$

$=\sqrt{2gh}$

$=\sqrt{2×9.8\mathrm{m}/{\mathrm{s}}^2×0.1\mathrm{m}}$

$=1.4\mathrm{m}/\mathrm{s}$

Now,

$H_1=H-h$

$=0.4m-0.1m$

$=0.3\mathrm{m}$

Kinematic equation

$y=y_0+v_{0}t+\frac{1}{2}a{t}^2$

$y-\frac{1}{2}a{t}^2$

(iv$y_0=0,v=0$and $\left.a=g\right\}$

$t^2=\frac{2H_1}{g}$

$t=\sqrt{\frac{2H_1}{g}}$

$=0.2474\mathrm{s}$

We know that

$v=\frac{x}{t}$

$x=Vt$

Therefore, the distance is $0.3464\mathrm{m}$.

If depth $h_2$for a second hole to be made to give the same value of $x$, we have,

$v_2=\sqrt{2g{h}_2}$

$0.3464\mathrm{m}/\mathrm{s}-\sqrt{2g{h}_2}×t_2$

$t_2=\frac{0.3464\mathrm{m}/\mathrm{s}}{\sqrt{2g{h}_2}}$

Vertical distance to be travelled by the water would be $\left(H-h_2\right)$, and it should be travelled in time $t_2$. Initial velocity would be zero in vertical direction. So, we have

$H-h_2=\frac{1}{2}g{t}_2^2$

$H-h_2=\frac{1}{2}g{\left(\frac{0.3464\mathrm{m}/\mathrm{s}}{\sqrt{2g{h}_2}}\right)}^2$

$0.4-h_2=\frac{1}{4}×\frac{0.12}{h_2}$

$1.6h_2-4h_2^2=0.12$

Solving this quadratic equation, we get

If depth $h_2$is too large, the time of the flight of the water would be smaller, so horizontal distance would be smaller.

If depth $h_2$is too small, horizontal component of the velocity would be smaller, so again horizontal distance $x$would be small. So, we need to have the value of $h_2$such that it balances the time and horizontal velocity. This can be achieved when $h_2=\frac{H}{2}$

$h_{max}=\frac{H}{2}$

Therefore, the depth for such a hole to be made to give maximum is $0.20\mathrm{m}$.